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Math Help - Mean Deviation.

  1. #1
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    Mean Deviation.

    I am doing some work with moment generating functions and have been given the question ''calculate the mean deviation'', this is defined to be E(lX-E(X)l).

    My m.g.f is (1-t/lambda)^-1 and was generated from f(x)=(lambda)e^-(lambda)x.

    I have no idea how to find MD from a m.g.f.

    cheers
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  2. #2
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    Quote Originally Posted by ardam View Post
    I am doing some work with moment generating functions and have been given the question ''calculate the mean deviation'', this is defined to be E(lX-E(X)l).

    My m.g.f is (1-t/lambda)^-1 and was generated from f(x)=(lambda)e^-(lambda)x.

    I have no idea how to find MD from a m.g.f.

    cheers
    Here is a blunt approach:

    Get E(X) = \frac{1}{\lambda} from the mgf. Now use the pdf to calculate E\left(\left |X - \frac{1}{\lambda} \right|\right).
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Here is a blunt approach:

    Get E(X) = \frac{1}{\lambda} from the mgf. Now use the pdf to calculate E\left(\left |X - \frac{1}{\lambda} \right|\right).
    Got E(X) = \frac{1}{\lambda} by differentiating m.g.f once then setting t=0.

    So do i manipulate the expectation of a random variable formula to look like:
    E\left(\left |X - \frac{1}{\lambda} \right|\right) = intergral |X - (1\lambda)| (lambda)e^-(lambda)x. dx?
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    Quote Originally Posted by ardam View Post
    Got E(X) = \frac{1}{\lambda} by differentiating m.g.f once then setting t=0.

    So do i manipulate the expectation of a random variable formula to look like:
    E\left(\left |X - \frac{1}{\lambda} \right|\right) = intergral |X - (1\lambda)| (lambda)e^-(lambda)x. dx?
    Yes. And note that \left |X - \frac{1}{\lambda} \right| = X - \frac{1}{\lambda} when X > \frac{1}{\lambda} and \left |X - \frac{1}{\lambda} \right| = \frac{1}{\lambda} - X when X < \frac{1}{\lambda}.
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